Detection Of Biological Molecules Essay, Research Paper

Detection of Biological Molecules

Introduction: Without C, N, H, S, O and P,

life wouldn & # 8217 ; t exist. These are the most abundant elements in life beings.

These elements are held together by covalent bonds, ionic bonds, H bonds,

and disulfide bonds. Covalent bonds are particularly strong, therefore, are present in

monomers, the edifice blocks of life. These monomers combine to do polymers,

which is a long concatenation of monomers strung together. Biological molecules can be

distinguished by their functional groups. For illustration, an amino group is

nowadays in amino acids, and a carboxyl group can ever be found in fatty acids.

The groups can be separated into two more classs, the polar, hydrophilic,

and the nonionic, hydrophobic. A fatty acid is nonionic, hence it doesn & # 8217 ; t blend

with H2O. Molecules of a certain category have similar chemical belongingss

because they have the same functional groups. A chemical trial that is sensitive

to these groups can be used to place molecules that are in that category. This

lab is broken down into four different subdivisions, the Benedict & # 8217 ; s trial for

cut downing sugars, the iodine trial for the presence of amylum, the Sudan III trial

for fatty acids, and the Biuret trial for amino groups present in proteins. The

last portion of this lab takes an unknown substance and by the four trials,

find what the substance is.

BENEDICT & # 8217 ; S Trial

Introduction: Monosaccharides and disaccharides can be detected because of

their free aldehyde groups, therefore, proving positive for the Benedict & # 8217 ; s trial.

Such sugars act as a reduction agent, and is called a reduction sugar. By blending

the sugar solution with the Benedict & # 8217 ; s solution and adding heat, an oxidation-

decrease reaction will happen. The sugar will oxidise, deriving an O, and

the Benedict & # 8217 ; s reagent will cut down, fring an O. If the ensuing solution

is ruddy orange, it tests positive, a alteration to green indicates a smaller sum

of cut downing sugar, and if it remains bluish, it tests negative.

Materials: onion juice5 trial tubes1 beaker murphy juice

rulerhot home base deionized waterpermanent

marker5 tongs glucose solutionlabels starch solution6 barrel

pipettes Benedict & # 8217 ; s reagent5 toothpicks

Procedure: 1.Marked 5 trial tubings at 1 centimeter and 3 centimeter from the underside. Label trial

tubes # 1- # 5. 2.Used 5 different barrel pipettes, added onion juice up

to the 1 cm grade of the first

trial tubing, murphy juice to the 1 cm grade of the 2nd, deionized H2O

up to the 1

cm grade of the 3rd, glucose solution to the 1 cm grade of the 4th,

and the

amylum solution to the 1 cm grade of the 5th trial tubing. 3.Used the

last barrel pipette, added Benedict & # 8217 ; s Reagent to the 3 cm grade of all 5

trial tubings and mix with a toothpick. 4.Heated all 5 tubings for 3

proceedingss in a boiling H2O bath, utilizing a beaker, H2O, and

a hot home base. 5.Removed the tubings utilizing tongs. Recorded colourss

on the undermentioned tabular array. 6.Cleaned out the 5 trial tubings with deionized

H2O.

Datas:

Benedict & # 8217 ; s Test Consequences

Discussion: From the consequences, the Benedict & # 8217 ; s trial was successful. Onion juice

contains glucose, and of class, glucose would prove positive. Starch doesn & # 8217 ; T

hold a free aldehyde group, and neither does potato juice, which contains amylum.

Water doesn & # 8217 ; Ts have glucose monomers in it, and was tested to do certain the terminal

consequence would be negative, a bluish colour.

IODINE Trial

Introduction: The iodine trial is used to separate amylum from

monosaccharoses, disaccharides, and other polyoses. Because of it & # 8217 ; s

unique coiled geometric constellation, it reacts with I to bring forth a blue-

black colour and trials positive. A xanthous brown colour indicates that the trial

is negative.

Materials: 6 barrel pipettespotato juicestarch solution 5 trial

tubeswateriodine solution onion juice

glucose solution5 toothpicks

Procedure: 1.Used 5 barrel pipettes, filled trial tubing # 1 with onion juice,

2nd with murphy

juice, 3rd with H2O, 4th with glucose solution, and fifth with

starch solution. 2.Added 3 beads of iodine solution with a barrel pipette,

to each trial tubing. Assorted

with 5 different toothpicks. 3.Observed reactions and recorded

in the tabular array below. Cleaned out the 5 trial tubings. Datas:

Iodine Test Consequences

Discussion: The iodine trial was successful. Potato juice and amylum were

the lone two substances incorporating amylum. Again, glucose and onion juice

contains glucose, while H2O doesn & # 8217 ; t incorporate amylum or glucose and was merely

tested to do certain the trial was done decently.

SUDAN III Trial

Introduction: Sudan III trial detects the hydrocarbon groups that are staying

in the molecule. Due to the fact that the hydrocarbon groups are nonionic, and

stick tightly together with their polar milieus, it is called a hydrophobic

interaction and this is the footing for the Sudan III trial. If the terminal consequence is

a seeable orange, it tests positive.

Material: scissorsdeionized watermargarineSudan

III solution petri dishstarchethyl intoxicant

forceps lead pencilcream5 barrel pipettes filter paper

cooking oilblow drier

Procedure: 1.Cut a piece of filter paper so it would suit into a petri dish. 2.

Used a lead pencil, and marked W for H2O, S for amylum, K for pick, C

for

cooking oil and M for oleo. Pull a little circle following to each

missive for the

solution to be placed. 3.Dissolve amylum, pick, cooking oil and

oleo in ethyl intoxicant. 4.Used a barrel pipette for each solution, added a

little bead from each solution to

the appropriate circled topographic point on the filter paper. 5.Allowed the

filter paper to dry wholly utilizing a blow drier. 6.Soaked the paper in the

Sudan III solution for 3 proceedingss. 7.Used forceps to take the paper from

the discoloration. 8.Marinated the paper in a H2O bath in the petri dish, changed

H2O often. 9.Examined the strength of orange discolorations of the 5 musca volitanss.

Record in the tabular array below. 10. Wholly dried the filter paper, and

washed the petri dish.

Datas: Sudan III Test Results

Filter paper:

Discussion: The consequences indicate that the Sudan III trial was sucessful. Water

and amylum decidedly doesn & # 8217 ; t incorporate any fatty substances. Cream and cookery

oil no doubtedly does incorporate lipoids. It was surprising to happen that oleo

doesn & # 8217 ; t incorporate any fat.

BIURET Trial

Introduction: In a peptide bond of a protein, the bond amino group is

sufficiently reactive to alter the Biuret reagent from bluish to purple. This

trial is based on the interaction between the Cu ions in the Biuret reagent

and the amino groups in the peptide bonds.

Materials: 6 trial tubesegg white solutionstarch

solution6 toothpicks rulerchicken soup

solutiongelatin6 parafilm sheets lasting

markerdeionized watersodium hydrated oxide labels

glucose solutioncopper sulphate

Procedures: 1.Used 6 trial tubings, and labeled them at 3cm and 5cm from the

underside. Labeled

each # 1 to # 6. 2.Added egg white solution to the 3cm grade of the

foremost tubing, chicken soup solution

to the 3-cm grade of the 2nd tubing, H2O to the 3 cm grade of the 3rd

trial tubing,

glucose solution to the 4th, amylum to the fifth, and gelatin to the

sixth, all at the

3 cm grade. 3.Added Na hydrated oxide to the 5 cm grade of each tubing and

mix with 6 different

toothpicks. 4.Added 5 beads of Biuret trial reagent, 1 % Cu sulphate,

to each tubing and mix

by puting a parafilm sheet over the trial tubing gap, and agitate

smartly. 5.Held the trial tubing against a white piece of paper, and recorded

the colourss and

consequences. Discarded the chemicals, and washed the trial tubing.

Datas:

Biuret Test Results

Discussion: The Biuret trial seemed to hold been successful. Glucose and amylum

are both saccharides, while H2O has no proteins. Egg white decidedly has

proteins, and so does gelatin. Chicken soup had a intimation of protein content.

Unknown Chemical # 143

Introduction: By executing the Benedict & # 8217 ; s Test, the Iodine Test, the Sudan III

Test, and the Biuret Test, chemical # 143 should be identified.

Materials: stuffs from the Benedict & # 8217 ; s Testmaterials from the Sudan

III Test Materials from the Iodine Testmaterials from the

Biuret Test

Procedures: 1.Performed the Benedict & # 8217 ; s Test, and recorded consequences. 2.

Performed the Iodine Test, and recorded consequences. 3.Performed the Sudan III

Test, and recorded consequences. 4.Performed the Biuret Test, and recorded consequences.

Datas: Properties of Chemical # 143

chemical # 143 was a white powderish substance.

Decision: After governing out the obvious incorrect substances from the list like

land java, egg white and yolk, table sugar and salt, sirup and honey, the

little sum of proteins was taken into factor. That besides eliminated powdered

skim milk, and soy flour. The low, or none fat content ruled out some more

picks like enriched flour. The lone picks left was maize amylum, glucose,

and murphy amylum. Because of the low reduction sugar, glucose can be ruled out

besides.

The amylum content of substance # 143 was really high. The protein content was

around the 10 % scope, so potato amylum would be a better conjecture so maize amylum.

But maize amylum contained merely a hint of fat when murphy amylum contained 0.8 % .

But 0.8 % is really undistinguished. The most educated conjecture to what chemical # 143

is potato amylum.